REVIEW ARTICLE
SIMPLE COMPUTATIONAL METHODS FOR THE ANALYSIS OF (S, S) TYPE
INVENTORY MODELS
Pandiyan, P., Palanivel, R.M., Kannadasan, K. and Vinoth, R.
Department of Statistics, Annamalai University, Annamalainagar, Tamilnadu, India
ARTICLE INFO ABSTRACT
A simple computational method (SCM) to analyze a class of (s, S) type inventory problem is developed. Under this (s, S) policy, (i) the number of units demanded where d = 1, 2… a ( s) at successive demand epochs form a Markov chain (MC) with one step transition probability matrix (TPM) P and (ii) the replenishments are instantaneous. This method gives the algorithm for computation of stationary probabilities of inventory process, joint probability function of number of transitions and quantities of replenishments per cycle, conditional and unconditional average costs. Illustrative example for a few special cases are provided, which strengthen the applicability of the SCM to practical problems.
INTRODUCTION
Numerous methods have been suggested by various authors for formulating and finding optimal inventory policies which have more number of practical applications in real life situations. Such eventualities with modifications have been increasing in the inventory policies of the (s,S) types also. One more possible way out is discussed below. In this paper, a different method of approach for the analysis of (s, S) inventory models, wherein the number of units demanded at successive demand epochs are Markov dependent as introduced by Krishnamoothy and Lakshmi (1991), is discussed. For the detailed discussion
*Corresponding author:[email protected]
about the systematic analysis of various types of (s, S) inventory system refer Srinivasan and Ravichandaran (1994), Hiller and Liberman, (1990) and Krishnamoorthy et al. (1995).
A Markov Dependent [(s, S), a, P] Inventory Model
For the present (s, S) inventory problem, it is assumed that the bulk quantity demanded, say ‘d’ at a demand epoch is a random variable with d = 1, 2, … , a ; a s such that the sequence of units demanded in the successive demand epochs forms a finite state Mc on the state space {1, 2, …, a } with TPM ‘P’. To avoid the perpetual shortage, it is assumed that (S-s) > s. Further if the inventory level is s+1 or greater at a demand epoch, then no order is placed. On the other hand, if the inventory
ISSN:
0975-833X
International Journal of Current Research Vol. 33, Issue, 3, pp.060-067, March, 2011
INTERNATIONAL JOURNAL OF CURRENT RESEARCH
Article History:
Received 13th December, 2010
Received in revised form 15th January, 2011
Accepted 7th February, 2011
Published online 13th March, 2011
© Copy Right, IJCR, 2011, Academic Journals. All rights reserved.
Key words:
level either falls to ‘s’ or dips below ‘s’ at an epoch, an order is placed for at least (S-s) units. The quantity ordered is subject to review at the epoch of replenishment so as to bring inventory to level S. Zero lead time is assumed. This kind of inventory model may be represented in symbols as [(s, S), a, P]. Further the inter occurrence times between successive demand epochs are assumed to be independent and identically distributed (i.i.d) random variables with distribution function G (.) and density function g (.). If the demand is for ‘i’ units at an epoch, it may be called as i-type demand for I = 1, 2… a.
Notations
{I (t)} : inventory level process. * : Convolution
E1 : {1, 2… a-1, a}
E 2 : {s+1, s+2… s-1, S}
E :
2 s a, , 2 S 1, a , , 2 s 3, , 2 s 2, , 2 s 1,
1 s a, , 1 S 1, a , , 1 s 3, , 1 s 2, , 1 s 1,
1,S-a
, 2,S-a2
, 3,S-a 1
, ,
a 1,S-a 1
,a -S a, , a -S 1, a , , a -S 3, , a -S 2, , a S 1,
2) S (2, 2) S (1,
3) S (2, 3) S (2, 3) S (1,
S) (a, ,
(3.S), S), (2, S), (1,
1) S (1,
E3 : {S-s, S-s+1 … S-s+ (a-1)}
E4 : {s-a+1, s-a+2… s-1, s}
E5(r) : {r, r+1… S-s}, where r is the smallest
integer
as S
N : {1, 2…}
Here, the set E is a proper subset of E1 E2 i.
e., E E1 E2. It is due to the fact that the
continuous time, inventory process decreases from the levels S, S-1, S-2 … etc, with instantaneous reflecting barriers at each of its levels i E4
depending upon the demands. Hence the inventory level never decreases to zero at any instant. Further E3 forms the set of all possible number of units of
replenishments. The set E5 (r) is the collection of
all possible number of transitions in a cycle of replenishments.
Markov renewal inventory process
Let 0 = T0 < T1 < T2 … be the successive demand
epochs and X0, X1 , X2, … be the number of units
demanded at these epochs respectively, so that the sequence X = { Xn } forms a MC on the states i,
j E1 with initial distribution p [ X0 = i ] = Pi, and
transition probabilities
X
j
X
i
P
i,
j
E
1P
n1 n ij (1)Assume that the TPM P = (Pi j) is irreducible and
a periodic throughout the discussions to follow. Let (S =) Y0, Y1, Y2 … represent the inventory
levels just after meeting the demands at T0, T1, T2,
…, so that time intervals between successive occurrences of bulk demands induce the triple sequence process { ( X, Y ), T } = { ( Xn , Yn ), Tn,
n N } as a Markov Renewal inventory process (MRIP) on the state space E with the Semi- Markov kernel Q (t) = [ {Q ( i, I ) , (j, J) ; t } ] where Q { (i. I), (j, J); t)}
=P
Xn1j,Yn1J
; Tn1Tnt
Xni,YnI
P(i,I), (j.J) G(t) 2)The TPM of the underlying (X, Y) double
sequence MC is given by Q () =
Q
((i, I), (j, J)) (say).Stationary Distribution of the (X, Y) MC
Let the stationary probability vector of the (X, Y) MC denoted by the row vector
π
π
i,j
for I = 1, 2 … a,j = s+1, s+2… S-a, i = 1,2, …,n, J = S-n ; n = a-1, a-2, …, 1
unique solution vector as the state space E is finite dimensional and P is irreducible. A Simple Computational Method (SCM) is capsule below as an algorithm.
STEP- I: Using the TPM P = (PiJ), I, J E1 obtain
its stationary probability vector, say = (1, 2 …
a) by solving p =, e = 1
STEP-II: Arrange the states as they are ordered in
the set E of section 1.1 and identify the TPM
Q
of the (X, Y) MC in terms of PiJ’s and zeros.STEP- III: Omit the last set ‘of ‘a’ equations corresponding to (J,S), J = 1,2, … a , from
Q Π
. Then select the remaining stationary equations together with
2
j
i j) (i, v
π
E
so that the
resulting independent set of simultaneous equations is solvable.
STEP- IV: Make suitable changes such that one set of ‘a’ equations corresponding to
1,S 1
,
π
2,S 2,
,
π
a,S aπ
appearing on theleft of the system
Π
Q
reduce to a system of ‘a’ equations of the form
π
,
π
,
,
π
A
.
1,S1 2,S2 a,Sa
Where Ais a full rank square matrix of order of ‘a’. Hence obtain values for
π
1,S1,
π
2,S2,
,
π
a,Saby any one of the standard methods.
STEP-V: Using the values of
1,S 1
,
π
2,S 2,
,
π
a,S aπ
and selectingexpressionsfor
1,S1,π2,S2, π2,Sa π2,Sa-1
π and
and compute the next set of ‘a’ components in the
solution of
Π
Q
. Continue it, until values for all componentsπ
i,j,
(i,
j)
E
of the II vectorare obtained.
Mean Sojourn times
For the MPIR {(X, Y), T}, let the mean sojourn time at state
π
i,j,
(i,
j)
E
be (i, J). Since theinter occurrence times between transition epochs are assumed to be i.i.d random variables, with d.f G (.), it is seen that
0 j
i, 1 G(t) dt μ
μ (4)
Inventory Level Process {I (t)}
Since I (t) gives the unhand stock level at time ‘t’, i.e.,
I
(t)
Y
nfor
T
n
t
T
n1 it is seen that {I (t) ; t 0 } is a semi -Markov Process on the state space E2 with Kernel Q Let
I
n
Y
i
.
P
(t)
0
Then by the application of key renewal theorem. It can be established thatP(n)
t)
i;
(n,
P
Lim
t
exists. Thus using theresults of (4), it can be shown that for i E1
a
1 j
n) (j, a
k) (1,
k) (1, k) (1, a
1 j
n) (j, n) (j,
π μ
π μ π P(n)
1 E
(5a)
where n = s+1, s+2,….,S-a and n = S
n -S
1 j
n) (j, a
k) (1,
k) (1, k) (1, n -S
1 j
n) (j, n) (j,
π μ
π μ π P(n)
1 E
(5b)
where n = S-a+1, S -a+2,….,S-1
Expected Total Costs
In this type of [(s,S), a, P] inventory situations, demand cum replenishment epochs play a very crucial role in obtaining cost expressions, Hence to study the characteristics of the time durations of the successive replenishment epochs Wn and Wn+1,
n N under steady state conditions. Assume that an i-type demand occurs at Wn while a J -type
demand takes place at Wn+1. Let is be called as
J-cycle. Further assume that ‘M’ denotes the quantity of replenishment at Wn+1, Z = (Wn+1- Wn) denotes
the random length of the cycle between the epochs Wn and Wn+1 and K denotes the random number of
demand epochs in it with the K-th one being the last demand epoch of the cycle i.e. at Wn+1.
Then materialization of the events (M = m), (K = k) and z (z, z + dz) implies that m E3, K E 5
(r) and z [0,). Hence for a given i, J E1,
(i) S1 = sum of the demands of the first
(k-1) demands in the J-cycle is strictly less than (S-s),
(ii) S2 = sum of the demands of all
k-demand epochs is equal to m and (iii) z Z z+ dz is to take the
following form
P ,P , ,P , ,g
z dzB 1 dz z k; m;
f *k
i i i i, i i, j
i, 1 2 k2 k
Where i, i1, i2, , ikE1, S1
Ss
S2 = m, m E3, K E 5 (6)
and B is the normalizing constant and ik = j. From
(6) with little effort one could obtain the marginal distribution for M, K and Z. The joint probability function (Pf) of M and K is sufficient to derive the cost per transition and is given by
, P , , P , P B 1 k) (m, f
1 1 k 2 1 i,i i ,j
i i,
j
(7)
i, i1, i2, …, ik E 1, s1 < S-s and S2 = m, m E 3
, K E 5 since iK = J. For the constructional details
of
f
j
.
refer appendices I (a), II (a), II (b), III (a)for some specific cases
Marginal Distributions in J-cycle
Let the marginal probability functions of quantity of replenishment M, number of transitions K in a J-cycle be respectively designated by
f
j(1)(m)
and(k)
f
(1)j
Thus
5
k
1 j
(1)
j (m) f(m,k); j
f
E
E
5
k
1 j
(2)
j (k) f(m,k); j
f
E
E
Further let
f
j* = probability that a cycle isterminated by a J-demand.
=
5 k 5
j m
k) (m, f
E E
Hence, the following are the conditional expected values of the random variables (m/K), K and M in the J-cycle.
f(m,k)
k m K
M E
3 5
E m kE
j
(8a)
5
k
(2) j j(K) k f (k) E
E
(8b)
3
m
(1) j j(K) m f (m) E
E
(8c)
Average Inventory per Transition
In a J-cycle, choose a demand epoch Tn at random,
Let (j, i) E be the state of the (Xn, Yn) process at
Tn. The inventory level during the transition
interval (Tn -Tn-1) is (j + i), if I = S and it is equal
to [S - M + j ] if i = S. Further let I+= I (
T
n) be the inventory level just after the demand epoch Tn.Hence the conditional expected inventory level during the interval (Tn-Tn-1), given that (Xn,Yn) =
(j,1) is given by
j
(j,S)i) , j S) (i
j(I ) (j i)π S E(M) jπ
E
2 i
E
( (9)
Cost Aspects
Consider the cost considerations for a j-cycle as Hj
the unit storage cost per transition Lj fixed ordering
cost Cj unit variable cost. Then the conditional
expected total cost per transition (TCT) in a j - cycle is given by
say , T ) (I E H K M E C (k) E
L (TCT)
E j j j j j
j j
j
(10a)
Thus the expected total cost per transition in any cycle is given by
1
j * j jf
T E(TCT)
E
Expected Long Run cost Per Unit time
For a given J E1, the expected long run total cost
(TC) per unit time in a j-cycle is given by
MZ
h E(I) EC (z) E
L (TC)
E j j j
j j
j
(11a)
Thus
a 1 j
j j(TVC)f
E
E(TC) (11b)
where
2 n
p(n) n E(TC)
E
0
3 5
m k
j
j (m / z) f (m,k;z) dz
Z M E
E E
and
0
3 5
m k j
j Z z f (m,k;z) dz
E
E E
OPTIMIZATION PROBLEMS
The best policies under two different contexts are worked out assuming specific values for the parameters involved. These are supported by numerical illustrations.
Example 1
It is assumed that a retailer keeps stock pile according to the [(S, s), a, P] system. The best policy given the values s = 2, a = 2, P11 = .3, P12 =
.4, L1= 150 L2= 200 C1 = C2 = 50, H1 = H2 = 10
when S = 5, 6 and 7 is determined by using the calculations of tables 1(a), 1(b) and 1(c) given below which are outlined in expressions (1) - (10) and appendices (I) through (III). Let P0 =
.6 .4
.7 .3 P0
The resulting calculation of E (TCT) for three values 5,6,7 of S is given in Table 1(c) which indicates that the over all minimum E (TCT) corresponds to S = 7. Hence [(2, 7), 2, P0] is the
most desirable policy, among the above three [(2, S), 2, P0]; S = 5, 6, 7 policies.
Example 2
It is assumed that the demand for the product is seasonal according to three different TPM’s obtained on the basis of the historical data. Now the problem is to identify the season which has the minimum E (TCT). For demonstration, [(2, 7), 2, P] policy is again investigated at three different seasons I, II and III respectively with P = P1
=
.6
.4
.7
.3
, P = P2 =
,
.5
.5
.5
.5
, P = P3
=
.6
.4
.1
.9
. Further, placing s =2, a = 2, L1 =
100, L2 = 150, C1 = C2 = 35, H1 = H2 = 8 the
corresponding results are computer, as reported in From this table, It may be observed that the season (*) corresponds to the P matrix which yields the highest v1 value. It may be due to the fact that the
number of transition in a cycle increases with increasing v1 values and this may minimize the
*j j j j
j j
j E(K) C E(M / K) H E(I ) f
L
E(TCT)
because number of transitions ‘K’ in a ‘j’ cycle appears in the denominator of the first two terms in the above E (TCT) expression.
APPENDIX I (a)
Joint PF of M and K for [(s, S), a, P] = [(2, 7), 2, P] model
5 k )P
P (P
4 k P P ) P (P
) P P P P (P ) P (P
3 k P
P )
P (P
B 1 k) (5, f
4 11 21 11
2 11 21 21 11
21 12 11 21 12 21 11
21 22 22
12
Table 1(a). ij values for [ (s, S), a, P] = [ (2,7), 2, P0] model
j 3 4 5 6 7
i .3 .4 .5 .6 .7 Ei (I+)
1 .0576 .0858 .0342 .1139 .0722 2.1525 2. .1372 .0797 .1889 0 .2306 3.5927 p(n) .1948 .1655 .2231 .1139 .3028 5.7452
jP(n) .5844 .6620 1.1155 .6834 2.1196 5.1649
Table 1(b).
f
j(m,
k)
Values for [(s, S), a, P] = [(2, 7), 2, P0] model3 4 5 fj(m) Ej(M) Ej(M/K) *
j
f
j=1 .1560 .0822 .0029 .2411 1.2055 .3657(k)
f
(2)1
.1560 .0822 .0029 .2411
(k)
k.f
(2)1
.468 .3288 .0145 .8113 = 1E (k)
j=2 .3290 .0221 0 .3511 4.2029 1.3028
.2340 .1672 .0066 .4078
(k)
f
2(2) .5630 .1893 .0066 .7589(k)
k.f
2(2) 1.689 .7572 .0330 2.4792 = E2 (k)Table 1(c). E (TCT) values for [(s, S), a, P] = [(2, S), 2, P0] models when S = 5, 6, 7
S J vj Lj/Ej(k) CjEj(M/k) HjEj(I) Tj * j
f
T
jf
j* E(TCT)5 1 .36 244.10 21.08 17.83 283.01 .29 82.07 242.72 2 .64 134,18 62.26 29.83 226.27 .71 1260.6
6 1 .36 265.02 12,21 19.45 296.68 .19 56.37 219.46 2 .64 96.36 72.69 32.29 201.34 .81 163.09
7* 1 .36 184.89 9.15 21.53 215.57 .24 51.74 165.09 2 .64 80.65 32.57 35.93 149.15 .76 113.35
Table 2. E (TCT) values for [(2, 7), a, P] = [(2, 7), 2, P] model when P varies
S J vj Lj/Ej(k) CjEj(M/k) HjEj(I) Tj *
j
f
T
jf
j* E(TCT) I 1 .36 123.26 12.80 17.22 153.28 .24 36.79 139.22 .64 60.49 45.60 28.74 134.83 .76 102.4
II 1 .50 78.03 16.59 22.99 117.69 .34 39.99 123.4 2 .50 65.81 37.24 23.47 126.52 .66 83.50
[image:6.504.105.405.195.359.2] [image:6.504.77.431.400.483.2] [image:6.504.73.446.532.606.2] 5 k 0 4 k P P P ) P (P 3 k P P ) P (P P P ) P (P B 1 k) (5,
f 11 21 11 11 12
22 12 21 11 12 21 22 12 2 4 12 11 21 22 12 22 12 11 12 21 12 21 11 22 22 12 2 P)P (P 4 k P P )P P (P ) P P P P P (P ) P (P 3 k P ) P (P B 1 k) (6, f where
5 3 k j 6 5 m 2 1 j k) (m, f 1 BAPPENDIX II (a)
Stationary Distribution of [(s, S), a P] = [(2, 6), 2, P] model STEP-I: ) P (P P ν and ) P (P P ν to leads 1 νe ν,
νP 1 21 12 21 2 12 12 21
STEP-: TPM of
Q
is found as toFrom (1, 3) (2, 3) (1, 4) (2, 4) (1, 5) (1, 6) (2, 6)
0 0 P P 0 0 0 0 0 P P 0 0 0 0 0 0 0 P P 0 P 0 0 0 0 0 P P 0 0 0 0 0 P P P 0 0 0 0 0 P P 0 0 0 0 0 (2,6) (1,6) (1,5) (2,4) (1,4) (2,3) (1,3) 21 22 11 12 11 12 22 21 12 11 22 21 12 11
STEP – III
P11 (1.4) + P21 (2.4) = (1.3)
(1)
P12 (1.5) = (2,.3)
(2)
P11 (1.5) = (1.4)
(3)
(1.3) + (1.4) + (1.5) + (1.6) = v1 (4)
(2.3) + (2.4) + (2.6) = v2 (5)
((1,6), (2,6)) P = ((1,5), (2,4)) (6)
STEP – IV
From (2) and (3), it is formed that
0 0 P P π , π π ,π(1,4) (2,3) (1,5) (2,4) 11 12
(7)
Addition of
π
(1,5),
π
(2,4)
P on both sides of (6),leads to
π
(1,5)
π
(1,6),
π
(2,4)
π
(2,6)
P
π
(1,5),
π
(2,4)
(I
P)
(8) Adding (8) with (7) P, and using the facts (4) and (5), it is obtained that
P
0 P 0 P P 0 0 P P P I π , π v 21 11 12 11 (2,4) (1,5)
APPENDIX II (b)
Joint PF of M and K for [(s, S), a, P] = [(2, 6), 2, P) model P P P ) P (P P P ) P (P P P ) P (P 0 B 1 k) (4, f 11 11 11 21 11 11 21 21 11 21 12 21 11 1 4 k 0 3 k P P ) P (P 2 k P )P P (P B 1 k) (5,
f 11 21 11 12
11 21 21 11 5 P P P P) (P P P ) P (P P P ) P (P 0 B 1 k) (5,
f2 12 22 21 12 11 21 12 22
where
4 2 k j 5 4 m 2 1 j k) (m, f 1 BAPPENDIX III (a)
Stationary Distribution of [(s, S), a P] = [(2, 5), 2, P] model STEP-I: ) P (P P ν and ) P (P P ν to leads 1 νe ν,
νP 1 21 12 21 2 12 12 21
STEP- : TPM of
Q
is found as To
0 0 0
0 0 0
0 0 0
0 0 0
21 22
11 12
12 11
22 21
12 11
P P
P P
P P
P P
P P 0 0 0
(2,5) (1,5) (1,4) (2,3) (1,3)
STEP - III: Selection of stationary equations
from
Q
Π
, e = 1 excepting the last set of‘2’ equations leads to
((1.5) , (2.5)) P = ((1.4) , (2.3))
(1)
P11(1.4) = (1,.3)
(2)
(1.3) + (1.4) + (1.5) = v1
(3)
((2.3) + (2.5)) = v2 (4)
Addition of P on both sides of (4) leads to ((1,5),
(1,4) (2.3) + (2.5)
P = ((1,4), (2,3)) (I + P) (5)
Rearrangement of (2) gives
0 0
0 P π , π ,0
π(1,3) (1,4) (2,3) 11 (6)
Addition (6) with (5) and using (3) and (4), it is simplified as
P
0 0
0 P P I π , π
vP (1,4) (2,3) 11
(7)
π
(1,4),
π
(2,3)
A
solving
(
7
);
it
it
is
seen
that
δ δ π and δ δ
π(1,4) 1 (2,3) 2
where A
and , A A
A A
22 21
12 11
1, 2, are the
determinant values with If
A A
A A δ , v v
A A δ , A A
v v δ
22 21
12 11
2 1
12 11 2 22
21 2 1
1
REFERENCES
Hiller, F.S and Liberman, G.J. 1990. Introduction to Operation Research, Fifth-edition, McCraw Hill.
Krishnamoorthy, A and Varghese, T.V. 1995. On a Two-commodity Inventory Problem with a Markov-Shift in demand, OPSEARCH, Vol.32, No.1, p44-56.
Krishnamoorthy, A and Lakshmi. B. 1991. An inventory model Markov dependent demand quantities, Cahiers,du, CERO Vol.33, No.1-2, 91-101.
Sivasamy. R and Senthamaraikannan. K 1994. Embedded processes of a Markov Renewal Bulk Service Queue, Asia Pacific Journal of Operation Research, 11-51.
Srinivasan. S.K and Ravichandran. N. 1994. Multi-iterm (s, S) inventory model with Poisson demand, General lead time and adjustable Recorder time. Proceedings of the international Conference on Stochastic Models, Optimization techniques and computer (p.226-237) applications, Willey Eastern Ltd.,,